$f(x)$ is continuous
$ \Rightarrow \mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} a + {\cos ^{ - 1}}\left( {x + b} \right) = f\left( x \right)$
$ \Rightarrow - 1 = a + {\cos ^{ - 1}}\left( {1 + b} \right)$
${\cos ^{ - 1}}\left( {1 + b} \right) = - 1 - a\,\,\,\,\,\,...\left( 1 \right)$
$f(x)$ is differentiate
$ \Rightarrow LHD = RHD$
$ \Rightarrow - 1 = \frac{{ - 1}}{{\sqrt {1 - {{\left( {1 + b} \right)}^2}} }}$
$ \Rightarrow 1 - {\left( {1 + b} \right)^2} = 1$
$ \Rightarrow b = - 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,....\left( 2 \right)$
From $\left( 1 \right) \Rightarrow {\cos ^{ - 1}}\left( 0 \right) = - 1 - a$
$\therefore - 1 - a = \frac{\pi }{2}$
$a = - 1 - \frac{\pi }{2}$
$a = \frac{{ - \pi - 2}}{2}\,\,\,\,\,\,...\left( 3 \right)$
$\therefore \frac{a}{b} = \frac{{\pi + 2}}{2}$
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$1.$ The correct statement$(s)$ is(are)
$(A)$ $f^{\prime}(1) < 0$
$(B)$ $f(2) < 0$
$(C)$ $f^{\prime}(x) \neq 0$ for any $x \in(1,3)$
$(D)$ $f^{\prime}(x)=0$ for some $x \in(1,3)$
$2.$ If $\int_1^3 x^2 F^{\prime}(x) d x=-12$ and $\int_1^3 x^3 F^{\prime \prime}(x) d x=40$, then the correct expression$(s)$ is(are)
$(A)$ $9 f^{\prime}(3)+f^{\prime}(1)-32=0$
$(B)$ $\int_1^3 f(x) d x=12$
$(C)$ $9 f^{\prime}(3)-f^{\prime}(1)+32=0$
$(D)$ $\int_1^3 f(x) d x=-12$
Give the answer question $1$ and $2.$